axis
axis
(h, k)
a.“c”
c =is1 the distance
b. V(2, 3from
)
focus
the vertex to the focus.
2 or y2
c.
F(
)
d.
x
2,
4
vertex
directrixGeneral form for
x2 opens
up or down
x2 parabola:
e.1 directrix _________
y
=
2
2 opens
2 + right
yare
or
left
yThere
=
(x
– h)
k
“c” units from
4c
f.
axis
_____________
x =vertex.
2
the directrix to the
The axis is the line that2 goes
y
=
¼(x
–
2)
+
3
g. equation:___________________
through the vertex and focus.
directrix
a. c = -3
b. V(-1, 2)
c. F(-1, -1)
d. x2 or y2
e. directrix _________
y=5
axis
f. axis
axis _____________
x = -1
y = -1/12(x + 1)2 + 2
g. equation:___________________
a. c = 2
(h, k)
b. V(-1, 4)
c. F( 1, 4 ) d. x2 or y2
General form for y2 parabola:
e.
_________
x = -3
1 directrix
x=
(y – k)2 + h
4c
f. axis _____________
y=4
axis
directrix
x = 1/8(y - 4)2 - 1
g. equation:___________________
directrix
axis
a. c =___
2
b. V(-4, 0)
c. F(-2,
0))
F(
d. x2 or y2
e. directrix x = -6
f. axis _____________
y=0
x = 1/8(y - 0)2 - 4
g. equation:___________________
axis
a. c =___
3
b. V(
V(1, 4))
c. F(1,
F( 7))
d. x2 or y2
e. directrix _________
y=1
directrix
f. axis _____________
x=1
y = 1/12(x - 1)2 + 4
g. equation:___________________
directrix
a. c =___
-3
b. V(0, -2)
c. F(
F(-3, -2)) d. x2 or y2
axis
e. directrix x = 3
f. axis _____________
y = -2
x = -1/12(y + 2)2 + 0
g. equation:___________________
Circles
General form:
(x - h
h)² + (y - k
k)² = rr²
Center (h, k)
radius = r
Using the form:
(x - h)² + (y - k)² = r²
Given: Center and radius
h k
2
Ex. 1: C(5,
r = 7
5 2)
(x - 5 )² + (y - 2 )² = 7²
(x - 5)² + (y - 2)² = 49
(x - h)² + (y - k)² = r²
h k
-3 4)
4
Ex. 2: C(-3,
r =2 5
(x - -3)² + (y - 4 )² = 2 5 ²
(x + 3)² + (y - 4)² = 20
(x - h)² + (y - k)² = r²
Given: Center & Another Point
h k
-7 & (5,
Ex. 3: C(4,
4 -7)
5 3)
3
(x - 4 )² + (y - -7)² = ²
(5 To
- 4)²
+ 7)²
find+r2(,3you
can= r²
(1)²
(10)²or= r²
plug in
the+ point
= r²
use the distance 101
formula
(x - 4)² + (y + 7)² = 101
(x - h)² + (y - k)² = r²
Ex. 4: C origin & (-5, 2)
(x - 0 )² + (y - 0 )² = ²
d  To
(x2find
 x1 )r ,you
(y2 can
 y1 )
22
2
2
2
plug
in
the
point
or
d  ( 5  0)  (2  0)
use the
formula
2 distance
2
d  ( 5)  (2)  25  4  29
(x - 0)² + (y - 0)² = 29²
x² + y² = 29
Given: Endpoints of diameter
Ex. 5: (2, 8) & (-4, 6) are
endpoints of the diameter.
2  the
4 8center
 6  using
1stFind
,
= (-1, 7)
C =

the
2
2 formula:
 midpoint

y

y
x

x


1
2
1
2
Let’s
use
Then…choose
either
,


2  like
 2 and finish
endpoint
C =(-1, 7) and (2, 8)
before.
h k
C =(-1,
-1 7)
7 and (2,
2 8)
8
(x --1 )² + (y - 7 )² = ²
(2 + 1)² + ( 8 - 7)² = r²
(3)² + (1)² = r²
10 = r²
(x + 1)² + (y - 7)² = 10
Remember how to
?!?
If a quadratic equation isn’t in
you will need to
to get it in the correct form.
Here’s how to do it:
x2 + y2 + 16x – 22y – 20 = 0
x2 + 16x + ( ) + y2 – 22y + ( )
=
20 +( ) + ( )
Rewrite the problem:
Group your x’s and leave a space.
Group your y’s and leave a space.
Move the constant and leave 2 spaces.
x2 + 16x +( 8)2 + y2 – 22y +( 11)2 = 20 +(64 ) +(121)
(x + 8)2 + (y – 11)2 = 205
Center (-8, 11) radius = 205
Complete the square
Half the linear term and square it.
Add to both sides.
Do this for both x and y.
Factor and simplify.
Now you try it:
x2 + y2 - 12x + 8y + 32 = 0
x2 - 12x +( 6) 2 + y2 + 8y +( 4)2 = -32 +(36) +(16)
(x - 6)2 + (y + 4)2 = 20
Center (6, -4) radius = 2 5
(x)² + (y)² = 36
Center (0, 0) radius = 6
up 6
Ex. 1:
left 6
right 6
Center (0, 0)
down 6
(x - 3)² + (y - 4)² = 25
Center (3, 4) radius = 5
up 5
Ex. 2:
down 5
left 5
right 5
Center (3, 4)
Ex. 3:
(x - 5)² + (y +4)² = 41
up 6.4
Center (5, -4) radius = 41 = 6.4
left 6.4
right 6.4
Center (5, -4)
down 6.4
2
x2
16
center: (0, 0)
+
2
y2
25
= 1
name of ellipse:
vertical
b: 4
a: 5
2, so you
b was under
the
x
Square
root
of
the
major
axis:
10
Square
root
of the
move b units
from
the
smaller
denominator.
focus (0, 3)
larger
denominator.
center
in
a
x
direction.
minor
axis:
2a
8
a was under the y2,
(0,
5),
(0,
-5),
center (0, 0)
so
you
move
a
units
vertices:
2b
focus (0, -3)
(4,
0),
(-4,
0)
from
the
center
in
c2 = a2 – b2
a foci:
y direction.
(0, 3), (0, -3)
2
x
+
9
center: (0, 0)
2
y
20
a: 2√5 b: 3
= 1
name of ellipse:
vertical
major axis: 4√5
minor axis: 6
vertices: (0, ±2√5)
(±3, 0)
foci: (0, ±√11)
center (0, 0)
__
x2 + __
y2 = 1
25
9
5
3
How Where
many units
is the
from
center
the center
to the
ofcurve
this ellipse?
in an “y”
“x” direction?
2
2
__
+
__
x
y = 1
16
36
4
6
How Where
many units
is the
from
center
the center
to the
ofcurve
this ellipse?
in an “y”
“x” direction?
2
2
2
2
x
+
y
x + 10y ==110
10
10
10
1
10
Divide to make the constant 1.
SF:
center: (0, 0)
vertices: (±√10, 0)
(0, ±1)
foci: (±3, 0)
2
x 2+
24x
72
3
2
2
y
+ 3y = =1 72
72
24
72
Divide to make the constant 1.
SF:
center: (0, 0)
vertices: (0, ±2√6)
(±√3, 0)
foci: (0, ±√21)
2
xx2
99
--
2
yy2
== 11
16
16
center: (0, 0)
a: 3 b: 4
“a” is“b”
theissquare
the square
root root of
vertices: (3, 0) (-3, 0)
of the
the
positive
negative
variable.
variable.
Will
go in
foci:
(5,the
0) direction
(-5, 0)
of the
positive
variable.
2
2
2
c =a +b
2
2
2
2 22
2
2
(y
+
1)
–
(x
–
3)
4(y
4(y ++ 1)
1) –– 25(x
25(x –– 3)
3)= ==1 100
100
25
100
4
100
100
Divide each term by 100 to get into form.
center: (3, -1)
a: 5 b: 2
vertices: (3, -6) (3, 4)
foci: (3, -1±√29)
Getting it into Standard Form
16x2 - 9y2 + 54y + 63 = 0
16x2 + (-9y2 + 54y + ( )) = -63 + ( )
Note:
Thethe
+54y
-6y
Factor
–9 becomes
out of the
“y” terms.
16x2 + -9(y2 - 6y + (32)) = -63 + -9( 9 )
Remember: Put
the –9 on 2the right too.
2
16x + -9(y - 3) = -144
-144
–144
-144
2
2
(y -3)
x =
1 by -144.
Divide– each
term
Why did the x and y
16
9 terms trade places?
(y
(y ––
16
16
2
3)
3)2
--
center: (0, 3)
a: 4 b: 3
vertices: (0, 7) (0, -1)
foci: (0, 8) (0, -2)
2
xx2
99
== 11
9x2 - 4y2 + 54x + 8y + 41 = 0
(9x2+54x+( ))+(-4y2+8y+( )) = -41+ ( ) + ( )
9(x2+6x+(32)) + -4(y2-2y+(12 )) = -41+ 9(9) + -4(1)
9(x + 3)2 – 4(y - 1)2 = 36
36
36
36
(x + 3)2 – (y – 1)2 = 1
4
9
(x +
4
2
3) 2
– (y –
9
center: (-3, 1)
a: 2 b: 3
vertices: (-5, 1) (-1, 1)
foci: (-3±√13, 1)
2
1) 2
= 1